Question

The line $\frac{x-1}{2}=\frac{y}{-1}=\frac{z+2}{2}$ cuts the plane $x+y+z=1$ at $P$. If the foot of the perpendicular from $P$ to a point $Q(3,-4,1)$ on the plane $S$ then the equation of the plane $S$ is

• A. $3x-2y-z=0$
• B. $2x-y+2z=12$
• C. $2x-10y+5z=51$
• D. none of these

Answer 1

The correct option is C $2x-10y+5z=51$

We have plne,

$x+y+z=\mathrm{1.......}\left(1\right)$

And line,

$\frac{x-1}{2}=\frac{y}{-1}=\frac{z+2}{2}=\lambda ......\left(2\right)$

Assume line (2) cuts plane (1) at point P,

$P=(2\lambda +1,-\lambda ,2\lambda -2)$

Since line (2) cuts the plane (1) , Therefore Point P lies on plane (1).

$2\lambda +1-\lambda +2\lambda -2=1$

$3\lambda =2$

$\lambda =\frac{2}{3}$

Therefore,

$P=(\frac{7}{3},\frac{-2}{3},\frac{-2}{3})$

direction ratios of normal of plane are,as we have two points, point P and Q(3,-4,1),

$(3-\frac{7}{3},-4+\frac{2}{3},1-\frac{-2}{3})$

$(\frac{2}{3},\frac{-10}{3},\frac{5}{3})$

Required Equation of plane,

$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$

$\frac{2}{3}(x-3)+\frac{-10}{3}(y+4)+\frac{5}{3}(z-1)=0$

$2x-10y+5z-51=0$

$2x-10y+5z=51$

Therefore option (C) is correct.